0-2-sin4x-1-sinx-cosx-dx-

Question Number 75848 by behi83417@gmail.com last updated on 18/Dec/19

\underset{0}{\int^{\frac{π}{2}}} \frac{\sin 4 x}{1 + sinx + cosx} dx = ?

Commented by mathmax by abdo last updated on 18/Dec/19

l e t I = \int_{0}^{\frac{π}{2}} \frac{s i n (4 x)}{1 + s i n x) c o s x} d x c h a n g e m e n t x = \frac{π}{2} - t g i v e

I = - \int_{0}^{\frac{π}{2}} \frac{s i n (2 π - 4 t)}{1 + c o s t + s i n t} (- d t) = \int_{0}^{\frac{π}{2}} \frac{- s i n (4 t)}{1 + s i n t) c o s t} d t = - I \Rightarrow

2 I = 0 \Rightarrow I = 0

Answered by MJS last updated on 18/Dec/19

\sin 4 x has a period of \frac{π}{2} \Rightarrow {we}^{'} re integrating

over a whole period \Rightarrow answer is 0

Commented by behi83417@gmail.com last updated on 19/Dec/19

e^{x} cellent dear proph .

than k s in advance sir .

Answered by MJS last updated on 19/Dec/19

\int \frac{\sin 4 x}{1 + \sin x + \cos x} d x =

= \int (\sin 3 x + \cos 3 x - 2 \cos 2 x - \sin x + \cos x) d x =

= \frac{1}{3} (- \cos 3 x + \sin x) - \sin 2 x + \cos x + \sin x + C

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